The perpendicular sides of a right angle isosceles triangle is 2acm .Find its hypotenuse?

Open in App

Solution

$Using Pythagoras theorem, Hypotenuse = \sqrt{{(Base)}^{2} + {(Height)}^{2}} = \sqrt{{(2 a)}^{2} + {(2 a)}^{2}} = \sqrt{8 a^{2}} ∴ Hypotenuse = 2 \sqrt{2} a cm$

Suggest Corrections

Similar questions

Q. One side other than the hypotenuse of a right-angled isosceles triangle is

4

cm. The length of the perpendicular on the hypotenuse from the opposite vertex is:

Q. What is the area of a right angled isosceles triangle whose hypotenuse is

6 \sqrt{2} c m ?

Q. In an Isosceles right angled

△ A B C

its hypotenuse

A C = 8

cm. Find the length of side

A B

and side

B C

Q. Assertion :If

x + k y = 1

and

x = a

are the equations of the hypotenuse and a side of a right angled isosceles triangle then

k = \pm a

. Reason: Each side of a right angled isosceles triangle makes an angle

\pm \frac{π}{4}

with the hypotenuse.

The sides forming the right angles and hypotenuse of a right angled triangle are xcm, (3x-2)cm and (2x+2)cm respectively. Calculate the length of the sides of the triangle.

Join BYJU'S Learning Program

The perpendicular sides of a right angle isosceles triangle is 2acm .Find its hypotenuse?

Using Pythagoras theorem,Hypotenuse=Base2+Height2=2a2+2a2=8a2∴Hypotenuse=22 a cm

$Using Pythagoras theorem, Hypotenuse = \sqrt{{(Base)}^{2} + {(Height)}^{2}} = \sqrt{{(2 a)}^{2} + {(2 a)}^{2}} = \sqrt{8 a^{2}} ∴ Hypotenuse = 2 \sqrt{2} a cm$